[proofplan] We verify four properties of $\Psi: L^2 \to \dot{H}^0$ in order: well-definedness (via [Plancherel](/theorems/247) and the compatibility of distributional and $L^2$ Fourier transforms), isometry (immediate from Plancherel), injectivity (the only polynomial in $L^2$ is zero), and surjectivity (construct a candidate via inverse Fourier transform and identify the Fourier-side difference as a distribution supported at $\{0\}$, hence a polynomial via the [Distributions Supported at a Point](/theorems/451) theorem). [/proofplan] [step:Verify that $\Psi$ maps $L^2$ into $\dot{H}^0$] Let $h \in L^2(\mathbb{R}^n)$. The [regular distribution](/page/Regular%20Distribution) $T_h \in \mathcal{S}'(\mathbb{R}^n)$ is a tempered distribution since $L^2(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n)$. Its equivalence class $[T_h] \in \mathcal{S}'(\mathbb{R}^n)/\mathcal{P}$ is well-defined. To verify $[T_h] \in \dot{H}^0(\mathbb{R}^n)$, we must show that $\widehat{T_h}\big|_{\mathcal{D}'(\mathbb{R}^n_0)}$ is a regular distribution whose $T$-representative satisfies $|\xi|^0\hat{f}_{T\text{-rep}} \in L^2(\mathbb{R}^n)$. For $h \in L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, both the pointwise integral definition and the distributional definition agree by the [Compatibility of the Distributional and Classical Fourier Transforms](/theorems/718). For general $h \in L^2(\mathbb{R}^n)$, choose $h_k \in L^1 \cap L^2$ with $h_k \to h$ in $L^2$. Then $T_{h_k} \to T_h$ in $\mathcal{S}'$, $\widehat{T_{h_k}} \to \widehat{T_h}$ in $\mathcal{S}'$, and $T_{\hat{h}_k} \to T_{\hat{h}}$ in $\mathcal{S}'$ (since $\hat{h}_k \to \hat{h}$ in $L^2$ by the [Plancherel Theorem](/theorems/247)). Since $\widehat{T_{h_k}} = T_{\hat{h}_k}$ for each $k$, the limit gives $\widehat{T_h} = T_{\hat{h}}$. It follows that $\widehat{T_h}\big|_{\mathcal{D}'(\mathbb{R}^n_0)}$ has $T$-representative $\hat{h}$. For $s = 0$, the membership condition $|\xi|^0\hat{h} = \hat{h} \in L^2(\mathbb{R}^n)$ holds by the [Plancherel Theorem](/theorems/247). Hence $[T_h] \in \dot{H}^0(\mathbb{R}^n)$. [/step] [step:Establish the isometry $\|[T_h]\|_{\dot{H}^0} = \|h\|_{L^2}$] \begin{align*} \|[T_h]\|_{\dot{H}^0} &= \||\xi|^0\,\hat{h}\|_{L^2} = \|\hat{h}\|_{L^2} = \|h\|_{L^2}, \end{align*} where the last equality is the [Plancherel Theorem](/theorems/247) (under the symmetric normalisation convention where the Fourier transform is a unitary isometry on $L^2$). [/step] [step:Prove injectivity by showing the only $L^2$ polynomial is zero] Suppose $\Psi(h_1) = \Psi(h_2)$, i.e., $[T_{h_1}] = [T_{h_2}]$ in $\mathcal{S}'(\mathbb{R}^n)/T_{\mathcal{P}}$. Then $T_{h_1} - T_{h_2} = T_p$ for some polynomial $p \in \mathcal{P}$. Suppose $p \neq 0$. Write $p(x) = \sum_{|\alpha| \le N} c_\alpha\,x^\alpha$ with $c_\beta \neq 0$ for some $|\beta| = N \ge 0$. For $|x|$ sufficiently large, $|p(x)| \ge \tfrac{1}{2}|c_\beta|\,|x|^N$ on a cone of positive solid angle. In polar coordinates $x = r\omega$ with $r = |x|$ and $\omega \in \mathbb{S}^{n-1}$: \begin{align*} \int_{\mathbb{R}^n} |p(x)|^2 \, d\mathcal{L}^n(x) &\ge \int_R^\infty \int_{\mathbb{S}^{n-1} \cap C} \frac{|c_\beta|^2}{4}\,r^{2N}\,r^{n-1} \, d\mathcal{H}^{n-1}(\omega) \, d\mathcal{L}^1(r) = \frac{|c_\beta|^2\,\sigma_C}{4}\int_R^\infty r^{2N+n-1} \, d\mathcal{L}^1(r) = \infty, \end{align*} where $C$ is the cone, $\sigma_C = \mathcal{H}^{n-1}(\mathbb{S}^{n-1} \cap C) > 0$, and $R > 0$ is large enough for the lower bound to hold. This contradicts $p \in L^2(\mathbb{R}^n)$. Therefore $p = 0$ and $h_1 = h_2$ in $L^2(\mathbb{R}^n)$. [/step] [step:Construct a preimage to prove surjectivity] Let $[f] \in \dot{H}^0(\mathbb{R}^n)$. By definition, $\hat{f}_{T\text{-rep}} \in L^2(\mathbb{R}^n)$. Define $h := \mathcal{F}^{-1}(\hat{f}_{T\text{-rep}}) \in L^2(\mathbb{R}^n)$, where $\mathcal{F}^{-1}$ is the $L^2$ inverse Fourier transform (which maps $L^2$ to $L^2$ by [Plancherel](/theorems/247)). We show $[T_h] = [f]$ in $\mathcal{S}'/T_{\mathcal{P}}$. By the well-definedness step, $\widehat{T_h} = T_{\hat{h}}$ in $\mathcal{S}'(\mathbb{R}^n)$, where $\hat{h} = \mathcal{F}(\mathcal{F}^{-1}(\hat{f}_{T\text{-rep}})) = \hat{f}_{T\text{-rep}}$. Therefore $\widehat{T_h}\big|_{\mathcal{D}'(\mathbb{R}^n_0)} = T_{\hat{f}_{T\text{-rep}}}\big|_{\mathcal{D}'(\mathbb{R}^n_0)} = \hat{f}\big|_{\mathcal{D}'(\mathbb{R}^n_0)}$. So the difference $\widehat{T_h} - \hat{f}$ vanishes on $C_c^\infty(\mathbb{R}^n_0)$, giving $\operatorname{supp}(\widehat{T_h} - \hat{f}) \subseteq \{0\}$. By the [Distributions Supported at a Point](/theorems/451) theorem, $\widehat{T_h} - \hat{f} = \sum_{|\alpha| \le N} c_\alpha\,\partial^\alpha\delta_0$. Set $g := T_h - f \in \mathcal{S}'(\mathbb{R}^n)$, so $\hat{g} = \sum_{|\alpha| \le N} c_\alpha\,\partial^\alpha\delta_0$. Each $\mathcal{F}^{-1}(\partial^\alpha\delta_0)$ acts as integration against a polynomial: $\mathcal{F}^{-1}(\partial^\alpha\delta_0) = (-i)^{|\alpha|}T_{x^\alpha/(2\pi)^n}$. Hence $g = T_p$ where $p(x) = \sum_{|\alpha| \le N} c_\alpha(-i)^{|\alpha|}x^\alpha/(2\pi)^n$ is a polynomial. Therefore $T_h - f = T_p \in T_{\mathcal{P}}$, so $[T_h] = [f]$ and $\Psi(h) = [f]$. [/step]